Data Sufficiency

Hi guys,

Need help with a small doubt. I have seen conflicting solutions on Kaplan.
Problem: (x^2 - 1)/(x+1) = 1
In such situations, it looks evident to cancel out the common term after we expand the numerator which gives us the result (x-1)=1 resulting in x = 2.
Agree!
However, I have another problem that shows that such canceling isn't a good idea.
(a^2- b^2) = (a-b)^2
Solution 1: By expanding both sides, we get (a+b)(a-b)=(a-b)(a-b), canceling out common terms i.e. (a-b), we get (a+b)=(a-b), further solving gives us b=0 (Unique value)

Solution 2: By expanding using formula on RHS gives us
a^2 - b^2= (a^2 -2ab + b^2)
Canceling out a^2, we get b= 0 or b= a (Not unique)

This is something I really am not sure about.
Plz suggest me whr am I missing out? Am i forgetting any rule here?? If so, how do I decide when to cancel and when not to cancel out like terms?

Awaiting your response.
Thank you!

Sorry did not get your ques....

Hi,
As you can see in the question, I mentioned 2 problems.
In the first problem- i cancel out the like terms and arrive at a unique solution. (here, canceling of like terms is accepted)

however, if u read my 2nd question: you can see 2 ways of solving that problem
1-- where I cancel out the like terms before expanding the formula on both sides
2-- where I expand and then cancel out the terms

Both result in 2 different solutions. Which is correct?

However, I have another problem that shows that such canceling isn't a good idea.
(a^2- b^2) = (a-b)^2
Solution 1: By expanding both sides, we get (a+b)(a-b)=(a-b)(a-b), canceling out common terms i.e. (a-b), we get (a+b)=(a-b), further solving gives us b=0 (Unique value)

First of all there is a difference between (a^2-b^2) = (a+b) (a-b)

2nd - (a-b)^2 = a^2-2ab+b^2

so both the formula's are different.....

Hey Varun,

Just to clarify from Prateek's statement, you're actually right that (x - y)^2 = (x - y)(x - y), so you're good there to expand to:

(a + b)(a - b) = (a - b)(a - b)

But the reason that you CANNOT simply divide both sides by (a - b) is that (a - b) could equal 0, which would make the equation:

(a + b)(0) = 0 * 0
and 0 = 0 so that does keep the equation intact

And more importantly, dividing by 0 is undefined...we just can't do it. So here you cannot simply cancel two algebraic terms if you do not know that they DO NOT equal 0.



Now, in your first case, you had:

(x^2 - 1)/(x + 1) = 1

here, there's no way that either the numerator or denominator could equal 0...otherwise the right hand side would not be 1, it would be 0 or undefined. So your first example forbids that the terms you're dividing could ever be 0, so you can freely divide there.

So, the rule of thumb is:

1) If the quantity that you're about to divide/cancel could equal 0, you can't simply divide...you must account for the possibility of 0.

2) If it's an inequality (> or <) and the quantity you're about to divide/cancel could be negative, you have to consider that possibility that that cancellation would flip the inequality sign.


3) If it's an equation (=) and the quantity that you're about to divide/cancel will not equal 0, then cancel away!

@Ian - brilliant minds think alike!

Geeeezzz.. yaaa. I almost forgot that.

Thanks Brian and Ian! Thank you for your quick reply--else the question would've eaten me up..lol

Appreciate your help...!

Hi Everyone,

Thanks for your explanations.

Varun - quick question about your original post. In Solution 2, where you expanded and then cancelled like terms, you stated that b=a and b=0. I get the b=a. How did you come up with b=0? Did you simply carry that over from the workout in Solution 1? And thanks to Ian and Brian we know that is not a possible solution.

Thanks again,
Michael

Hi Kuan, Sorry for a late reply.

Solution 2: By expanding using formula on RHS gives us
a^2 - b^2= (a^2 -2ab + b^2)
Canceling out a^2, we get b= 0 or b= a (Not unique)

I got b=0 by canceling terms on both sides:
a^2 - b^2= (a^2 -2ab + b^2)
- b^2= -2ab + b^2 (canceling a^2)
Now, bringing b^2 terms on one side, we get -2(b^2) = -2ab
or b^2 = ab
here, 2 cases can be considered, when b=0 and when it isn't.
when b=0 we get 0=0
when b not= 0, we can cancel b and get b=a.

Hope it is clear now.